Redshift Academy


Complex Scalar Fields
---------------------

Think of 2 real scalar fields φ₁ and φ₂ and combine them together
in the following manner.

φ = φ₁ + iφ₂ and φ^* = φ₁ - iφ₂
 
We can write that ∂φ/∂x^μ = ∂φ₁/∂x^μ + i∂φ₂/∂x^μ
    .
L = φ₁²/2 - (∂_xφ₁/2)² - (∂_yφ₁/2)² - (∂_zφ₁/2)² - m²φ₁²
         .
       + φ₂²/2 - (∂_xφ₂/2)² - (∂_yφ₂/2)² - (∂_zφ₂/2)² -  m²φ₂²

This can be simplified to:
    ..               
  = φφ^*/2 - ∂_xφ^*∂_xφ - ∂_yφ^*∂_yφ - ∂_zφ^*∂_xφ -  m²φ^*φ/2                 

It is conventional to drop the 1/2.  Therefore, in abbreviated form, 
we can write:

L = ∂_μφ∂^μφ^* - m²φ^*φ

The quantity m²φ^*φ is referred to as the 'mass term'.

Now apply a phase transformation, exp(iθ), to φ. This can be considered 
as a 'rigid' or global transformation that changes all points in the  
field by the same amount.  It is easy to see that the Lagrangian is
invariant under this transformation (φ' = φexp(iθ) and φ'^* = φ^*exp(-iθ)).
This implies there is a symmetry and a conserved quantity.  

Now e^iε = 1 + iε   ... Taylor series

Thus,

φ -> φ + iεφ -> φ + εf_φ where f_φ = iφ

φ^* -> φ^* - iεφ^* -> φ^* - εf_φ^* where f_φ^* = -iφ^*

Look for a conserved quantity using Noether's Theorem. 
         .   .
π_φ = ∂L/∂φ = φ^*
         .    .
π_φ^* = ∂L/∂φ^* = φ

Conserved quantity = ∫[π_φf_φ + π_φ^*f_φ^*]d³x
                        .     .
                   = i∫[φ^*φ - φφ^*]d³x

The quantity in [] is the charge density of the field, ρ.  

Another way to write this is:
        .
ρ =  Im(φφ^*)

∴ Q = ∫ρd³x

We can also write a 4-vector for current, j^μ, as:
   
j^μ = Im(φ^*∂^μφ)

The word 'charge' can have several different meanings.  It can be the
electric charge but also the weak hypercharge.  The take away from all of
this is that the rotational symmetry of a field is associated with 
charge.  THUS, THE U(1) SYMMETRY IS ASSOCIATED WITH CHARGE.  

Notes:

 - Every field has a field quanta. E-M field => photon.
 - Every particle is a quanta of a field.
 - CHARGED PARTICLES ARE ALWAYS DESCRIBED BY COMPLEX FIELDS. 
 - Don't confuse ρ with the e-m field.  They are different things. 
 - Need to couple ρ with e-m field in a Lagrangian to see how they 
   interact.  

Gauge Theory
------------

A symmetry is a transformation that preserves relevant physical properties
such as the equations of motion.  It can be described as a transformation 
that leaves the Lagrangian invariant.  A GLOBAL symmetry is a symmetry 
transformation that is executed by the same 'amount' at every point. A LOCAL
symmetry has a transformation that depends on space-time. In the case of 
rotation, the angle of the rotation can change for different locations and times

A gauge symmetry is a local symmetry that relates states that are physically 
the same. The states it connects are then only different in their mathematical 
structure. The presence of a gauge symmetry can be viewed as a redundancy 
in description.  In the words there is more description and more mathematical
structure than is strictly necessary to describe the physical state.  Gauge
transformations represent a stronger symmetry than the global symmetry 
associated with a rigid transformation where θ is not a function of position.  
We will see how the use of a gauge transformation can make an otherwise 
invariant Lagrangian invariant.

Consider a complex field φ = φ_R + iφ_I and  φ^* = φ_R - iφ_I and rotate by a phase 
θ(x) to get φ' as follows:

φ' = φe^iθ(x) and φ'^* = φ^*e^-iθ(x)  

This is a local U(1) symmetry.

Recall from the previous section on complex fields that:

L = ∂_μφ'∂^μφ'^* - m²φ'φ'^*

After some math the Lagrangian density becomes (dropping the factor of 1/2 to 
simplify matters):

(∂_μφ - i∂_μθφ)(∂_μφ^* + i∂_μθφ^*) - m²φφ^*

The m²φφ'^* term is invariant but the rest is not.  In order to make the 
Lagrangian invariant under the local U(1) symmetry, we need to add the 
concept of the GAUGE COVARIANT DERIVATIVE, D_μ.  For the electromagnetic 
field D_μ is defined as:

D_μφ = ∂_μφ + iqA_μφ and 

D^μφ^* = ∂^μφ^* - iqA^μφ^*  

where A_μ is the GAUGE FIELD written as a 4-vector = (A₀,A₁,A₂,A₃).  In the 
case of the electromagnetic field A₀, is the electrostatic potential and
A_1..3 is the magnetic vector potential E = -∇A₀ - ∂A_1..3/∂τ and B = ∇xA_1..3.  
Note: A_μ itself is not itself gauge invariant.  q is the charge (coupling 
constant).  The electric charge is just a measure of the strength of a 
particle's interaction with the electromagnetic field. Saying that a 
particle has a given charge is the same as saying it interacts with the
electromagnetic field with a certain strength. 

In order for the Lagrangian density to be invariant it is necessary that 
A transform as:

A_μ' = A_μ - (1/q)∂_μθ

Proof:

D_μφ' = ∂_μφ' + iqA_μ'φ' 

     = (∂_μφ + iφ∂θ/∂x)e^iθ(x) + iq(A_μφ - (1/q)φ∂θ/∂x)e^iθ(x)

     = (∂_μφ + iqA_μφ)e^iθ(x)

     = e^iθ(x)D_μφ

Similarly,

D^μφ'^* = e^-iθ(x)D^μφ^*

Therefore,

D_μφ'D^μφ'^* = D_μφD^μφ^* and the Lagrangian is gauge invariant.

We can rewrite the Lagrangian as:

L = D_μφD^μφ^* - m²φφ^*

  = (∂_μφ + iqA_μφ)(∂_μφ^* - iqA^μφ^*)e^iθ(x)e^-iθ(x) - m²φφ^*

  = ∂_μφ∂^μφ^* + q²A_μA^μφφ^* + iqA_μ[φ∂^μφ^* - φ^*∂^μφ] - m²φφ^*

This Langrangian is invariant under the local U(1) symmetry. 
The [] term represents a current and A_μ[] represents the 
interaction between this current with the electromagnetic 
field.  Thus, we see that our theory which is invariant under 
local gauge transformations is promoted to an interacting theory. 

Invariance of A_μ by Itself
--------------------------

We have looked at how the introduction of the gauge field makes
φ invariant under a local transformation. However, the gauge
field itself must also have its own gauge invariant dynamic term 
in the Lagrangian.  Consider the construction, F_μν, defined as:

F_μν = ∂_μA_ν - ∂_νA_μ

Intuitively, this is a good choice since it contains the first 
derivatives, ∂_μ, of the field that is characterisitic of all 
Lagrangians.  

If we plug in A_μ' = A_μ - (1/q)∂_μθ into the above we get:

F_μν = ∂_μA_ν + ∂_μ∂_νθ - (∂_νA_μ + ∂_ν∂_μθ)

  = ∂_μA_ν - ∂_νA_μ 

Thus, under the gauge transformation, F_μν, remains unchanged.  To make 
this both gauge and Lorentz invariant and also match the quadratic
nature of the other terms in the Lagrangian, it is necessary to form 
the product:

F_μνF^μν

We can now rewite our a Lagrangian for the gauge field as:

L = -F_μνF^μν

The is the Lagrangian that describes that electromagnetic field in 
the absence of any charges (complex fields are charged fields so in
the absence of charge φ = 0).

The Higgs Mechanism
--------------------

In the previous section we have talked about the role of gauge fields 
in constucting invariant Lagrangians.  The quanta associated with 
these fields are called GAUGE BOSONS.  By analogy with the m²φφ* term 
associated with the complex field, we might expect that mass for the 
gauge bosons might be associated with a m²A_μA^μ term.  However, such 
terms are not invariant and therefore break the local symmetry.  The 
invariance can be seen since m²(A_μ + ∂_μθ)(A^μ + ∂^μθ) is clearly not equal 
to m²A_μA^μ.  The q²A_μA^μφφ^* term from before represents an interaction term 
and IS invariant because A_μ and A^μ have already been transformed during
the derivation.  The consequence of this invariance is that, in general, 
we can make the statement that gauge symmetry prohibits gauge bosons 
from having mass.  

The standard model is a gauge theory and particles do have mass.  The 
implication of this is that for mass to appear in the standard model, 
the local symmetry has to be broken.  This is referred to as SPONTANEOUS 
SYMMETRY BREAKING.

Bosonic Mass
------------

Nearly all of the mass of protons and neutrons is from the kinetic energy 
of their constituent quarks and gluons.  This kinetic energy can only exist
if the quarks themselves have a rest mass.  It is the Higgs field that is
responsible for giving mass to these constituent particles and also to the
electrons.  The Higgs field is also responsible for giving mass to the REAL
W⁺, W^- and Z⁰ gauge bosons that can be produced in collisions if sufficient 
energy is available.




The potential energy, V(φ), of the Higgs field has the shape of a 
'Mexican hat'.  For ordinary fields the lowest energy is the point 
corresponding to φ = 0.  For the Higgs, however, the location of the
lowest energy is not at φ = 0 but at the point φ = f.

Any system, including a universe, will tumble into its lowest energy state, 
like a ball bouncing down to the bottom of a valley.  The universe, like a 
ball, comes to rest somewhere on this circular trench, which corresponds 
to a nonzero value of the field.  That is, in its natural, lowest energy 
state, the universe is permeated throughout by a nonzero Higgs field. 

The Higgs field is a complex field represented by:

φ = φ_R + iφ_I and  φ^* = φ_R - iφ_I

The equation representing the potential of the Higgs field is of the form:

V(φ) = -m²φ² + λφ⁴ where λ is a coupling constant.

Note:  +m²φ² is a parabaloid centered at 0.

For the minima:

∴ dV(φ)/dφ = -2m²φ + 4λφ³ = 0

so that,

φ = f = m/√2λ

f is referred to as the VACUUM EXPECTATION VALUE.  The VEV for the
Higgs field is 246 GeV.

The Higgs potential can also be written as:

V(φ) = -m²φφ^*+ λ(φφ^*)²

In this system, polar coordinates are much easier to work with than 
cartesian coordinates.   We can write:

φ = ρe^iα and φ^* = ρe^-iα 

Where ρ represents the radial coordinate and α represents the
angular coordinate.

Abelian U(1) Gauge Theory 
-------------------------

The particle at point A is symmetrical about rotation of the complex plane.  
Thus, the Higgs field has U(1) symmetry.  When this symmetry is spontaneously 
broken, the particle 'descends' into the trough (point B) where the energy 
is at a minimum.  At this point the particle experiences a flat potential as 
it moves along a circular path as shown.  A movement in this direction does 
not face any resistance since the energy in the adjacent state is the same.  
A particle moving in this angular fashion therefore has zero mass.  This is 
the GOLDSTONE boson*.   In contrast, oscillations back and forth along the 
radial (red) direction are not massless.  These are the HIGGS boson excitations.  
Mathematically, we can get some insight into the Higgs mechanism as follows:

*Goldstone's Theorem:  For  each  broken  generator  of  the  original symmetry  
group one massless spin-zero particle will appear.

We can write the Lagrangian for a spin 0 field (the Higgs boson is spin 0 
particle) as:

L = ∂_μφ∂^μφ^* - V(φ) 

  = (∂_μφ_R)² + (∂_μφ_I)² - V(φ) 

In terms of ρ and α this is equivalent to:

L = (∂_μρ)² + ρ²(∂_μα)² - V(ρ) 

Now ρ = f and is fixed and V(ρ) -> 0.  Therefore, the Lagrangian can 
be written as:

L = 0 + f²(∂_μα)² - 0 

  = f²(∂_μα)² 

  = (∂_μfα)²

This is characteristic of a field with quanta of zero mass.  This is 
the GOLDSTONE FIELD.

ρ represents the distance to the bottom of the trough, f, plus a 
small displacement, H.  Thus we can write:

ρ = f + H

The amount of energy required to excite the Higgs is quite large.  
If we consider a low energy where the oscillations are small we 
can write:

φ = fe^iα

Under a local U(1) transformation the field transforms as:

φ -> φexp(iθ(x)) 

However, this is not symmetric since ∂_μφ∂^μφ^* ≠ ∂_μφ'∂^μφ'^*.  In order 
to make it so we need to introduce the GAUGE FIELD, A_μ where A_μ 
transforms as  A_μ -> A_μ + ∂_μθ and use the covariant derivative.
This results in:

D_μφ = ∂_μφ + iqA_μφ

    = i(∂_μα + qA_μ)fe^iα

Likewise, 

D^μφ^* = -i(∂_μα + qA_μ)fe^-iα 

Therefore,

D_μφD^μφ^* = (qA_μ + ∂_μα)²f²

     = q²f²A_μA^μ + 2qf²A_μ∂_μα + f²(∂_μα)²

If we now make the UNITARY gauge transformation A_μ -> A_μ - (1/q)∂_μα we get:

D_μφD^μφ^* = q²f²[A_μ - (1/q)∂_μα][A^μ - (1/q)∂_μα] + 2qf²[A_μ - (1/q)∂_μα]∂_μα + f²(∂_μα)²

     = q²f²A_μA^μ - 2qf²A^μ∂_μα + f²(∂_μα)² + 2qf²A_μ∂_μα - 2f²(∂_μα)² + f²(∂_μα)²
   
     = q²f²A_μA^μ

     ≡ m_γ²A_μA^μ

So,'mathemagically' the Goldstone boson disappears.  This explains 
why it is never seen in nature.  In the process of disappearing, the 
Goldstone boson has given rise to a mass m²A² equivalent to the mass  
of a photon.  It is instructive to count the degrees of freedom before 
and after SSB has occurred.  We started out with a massless photon 
with 2 degrees of freedom and a complex scalar field with 2 degrees
of freedom for a total number of 4 degrees of freedom.  After the 
SSB we have one massive photon with 3 degrees of freedom and a real 
scalar field with 1 degree of freedom, again for a total of 4 degrees
of freedom.

So this is a little confusing since we know the 'everyday' photon is
massless.  We must conclude that spontaneous symmetry breaking does not 
occur for the photon, or does it?.  There is a special case where 
spontaneous symmetry breaking does occur and that is within a 
superconductor.  In the theory of superconductivity the photon does 
in fact have a mass.

So far we have assumed that the H in f + H is small and oscillations 
of the field can be ignored. However, if the system is given enough 
energy, the field will start oscillating in the radial direction along 
ρ (the red curve).  These oscillations represent the HIGGS BOSON.

As before, we can choose a direction of fluctuation so that the vacuum 
Higgs field is:

φ = (f + H)e^iα

The Lagrangian now looks like:

L = D_μ{(f + H)e^iα}D^μ{(f + H)^*e^-iα} - m²(f + H)² - λ(f + H)⁴ - F_μνF^μν

  = (∂_μ + iqA_μ)(fe^iα + He^iα)(∂^μ - iqA^μ)(fe^-iα + He^-iα) - m²(f + H)²  

     - λ(f + H)⁴ - F_μνF^μν

After some math The Lagrangian becomes:

L = q²f²A_μA^μ + 2qf²A_μ∂_μα + f²(∂_μα)² + ∂_μH∂^μH - 4λf²H² - F_μνF^μν 

    + cubic and quartic terms

This Lagrangian now describes a theory with a photon of mass m_γ = qf, 
a Higgs boson, H, with m_H = 2√(λ)f and a massless Goldstone boson, α.  

Again, we can make the UNITARY gauge transformation A_μ -> A_μ - (1/q)∂_μα 
to get:

L = q²f²A_μA^μ + ∂_μH∂^μH - 4λf²H² - F_μνF^μν + cubic and quartic terms

Now there are no massless particles in this theory, and the α field has 
completely disappeared from the Lagrangian.

Thus, when the Higgs field is excited, the Higgs boson also get its 
mass as a consequence of spontaneous symmetry breaking.  

Note:  It is inportant to distinguish between the Higgs field and the
Higgs boson. The Higgs field is the entity that gives all other particles 
a mass.  The Higgs boson is a particle (quanta of the field) that gets 
its mass like all other particles.  Once the Higgs particle has been 
created, it will eventually decay.  Though the Higgs particle interacts 
with all massive particles it prefers to interact with the heaviest 
elementary particles.  The Higgs particle is considered to be a carrier 
of a force.  It is a boson, like the other force-transferring particles: 
photons, gluons and electroweak bosons.  It has 0 spin, carries no
charge and has a mass of about 126 GeV.

Non-Abelian SU(2)⊗U(1) Gauge Theory (Yang-Mills)
------------------------------------------------

In the interest of simplicity, the above arguments have only considered
the Abelian U(1) gauge symmetry breaking and are by no means rigorous.  
The discussion has been focussed around the photon and the mechanism for 
gaining mass.  The sole purpose has been only to provide a basic idea of 
the fundamental concepts and in this respect the Abelian U(1) model is 
sufficient.

For the SU(2) symmetry that occurs in the electroweak theory of the 
Standard Model, the elements in the Lagrangian are a bit different. 
The gauge symmetric theory for SU(2) was developed by YANG and 
MILLS.  Under a local SU(2) transformation the field transforms as:

φ -> φexp(iθ_i(x)σ_i/2)

where σ_i are the Pauli matrices - the generators of SU(2).  Note: 
These are NOT the spin angular momentum generators.  They are, instead, 
asscociated with the weak isospin.  The physical interpretation is
completely different, but the algebra is identical to that of angular 
momentum.  θ_i are the 3 group parameters.

The covariant derivative for a SU(2)⊗U(1) gauge invariant Lagrangian 
becomes:

D_μ = ∂_μ + igW_μⁱσ_i/2 + ig'Y_WB_μ

Where g and g' are coupling constants.

The W_μⁱ fields are associated with the weak isospin, the generator of 
SU(2).  Since there are 3 generators, there are 3 gauge fields (with 
massless bosons) designated by W_μ⁰, W_μ¹ and W_μ².  Likewise the B_μ field 
is associated with the weak hypercharge, Y_W, the generator of U(1).  Note 
that this U(1) is NOT the same as the U(1) in the Abelian case discussed 
above for the photon.  In that case, U(1) is associated with the 
conventional charge, q.  By analogy with photons, particles with 
weak hypercharge interact by the exchange of B bosons - the massless 
quanta of the B gauge field. 

Due to the non-commutativity of the generators of this symmetry 
([σ₁,σ₂] = iσ₃ etc.), the gauge tensor for the W fields is:

G_μνⁱ = ∂_μW_νⁱ - ∂_νW_μⁱ + f^ijkW_μ^jW_ν^k

where f^ijk is the structure constant of the particular gauge group.

The basic form of the Langrangian that encapsulates SU(2)⊗U(1) looks
like (omitting constants):

L = D_μφD^μφ^† - V(φ) - F_μνF^μν - G_μνⁱG^μν_i

where F_μν has been redefined as:

F_μν = ∂_μB_ν - ∂_νB_μ

In the Abelian case the Higgs field was considered to be a complex 
scalar field with 2 components.  In the non-Abelian case the Higgs 
field is a complex scalar field of the group SU(2) with 4 components 
- 2 neutral (0) and 2 charged (Q) each with 1 degree of freedom for 
a total of 4.  Neglecting normalization factors, the field is a 
weak isospin doublet written as:

     -  -     -       -     -        -
φ = | φ⁺ | = | φ₁ + iφ₂ | ≡ | H⁺ + iH^- | with I₃ = 1/2 and Y_W = 1/2 
    | φ⁰ |   | φ₃ + iφ₄ |   | H + iH ⁰| with I₃ = -1/2 and Y_W = 1/2 
     -  -     -       -     -        -

and its Hermitian conjugate,
      -     -
φ^† = | φ^- φ^0* |  
      -     -

Under U(1) rotations, these doublets are multiplied by a phase exp(iφY_W)

Gauge symmetry is spontaneously broken down to the U(1) symmetry of 
electromagnetism.  This can be written as:

SU(2)_L⊗U(1)_Y -> U(1)_EM

Where L denotes left-handed chirality, Y_W denotes the weak hypercharge 
and EM denotes the electromagnetic field. 

As in the Abelian case, for low energies (H very small) we can write: 

D_μφD^μφ^† = (∂_μφ + igW_μⁱσ_iφ/2 + ig'Y_WB_μφ)(∂_μφ^† - igW_μⁱσ_iφ^†/2 - ig'Y_WB_μφ^†)

After some math and, for the purpose of illustration, neglecting 
all constants, we get:

D_μφD^μφ^† = f²[g²{(W_μ¹)² + (W_μ²)²} + (gW_μ⁰ - g'B_μ)²]

The g² term can be written as W_μW_μ^† where W_μ and W_μ^† are LADDER 
operators defined as:

W_μ = W_μ¹ + iW_μ² = W^- 

W_μ^† = W_μ¹ - iW_μ² = W⁺ 

Ladder operators were introduced in the context of angular momentum 
and rotation so it is also appropriate to use them in the context of 
rotations associated with SU(2).  

Thus, we can write:
 
f²g²{(W_μ¹)² + (W_μ²)²} = f²g²W_μW_μ^† and identify the mass m_W = gf

The 2 remaining neutral Z and γ bosons are defined in terms of the 
WEINBERG ANGLE, θ_W, as follows:

Z_μ⁰ = W_μ⁰cosθ_W - B_μsinθ_W

γ_μ = W_μ⁰sinθ_W + B_μcosθ_W

θ_W is found experimentally and is the angle by which SSB rotates the 
original W_μ⁰ and B_μ vector boson plane.  It is defined as:

cosθ_W = g/√(g² + g'²)

sinθ_W = g'/√(g² + g'²)

It also gives the relationship between the masses of the W and Z 
bosons as follows:

m_Z = m_W/cosθ_W

Therefore,

Z_μ⁰ = gW_μ⁰/√(g² + g'²) - g'B_μ/√(g² + g'²)
   
  = (1/√(g² + g'²))(gW_μ⁰ - g'B_μ) 

with mass, m_Z = m_W/g/√(g² + g'²) = gf/g/√(g² + g'²) = f√(g² + g'²) 

and

γ_μ = g'W_μ⁰/√(g² + g'²) + gB_μ/√(g² + g'²)

  = (1/√(g² + g'²))(g'W_μ⁰ + gB_μ) 

with mass m_γ = 0

It is instructive to look at the degrees of freedom before and after SSB 
has occurred.  At the outset there are 4 degrees of freedom associated
with the Higgs field, 2 associated with the massless B field and 6 
associated with the 3 massless W fields for a total of 12.  After SSB 
there is a real scalar field H with 1 degree of freedom, 3 massive weak 
bosons, W^± and Z⁰, with 9, and 1 massless photon with 2, again yielding 
a total of 12.  The scalar field degrees of freedom have been 'eaten' 
to give the W^± and Z⁰ bosons their longitudinal components.  Note: 
massless particles cannot have a degree of freedom in their longitudinal 
direction since they are constrained to travel at the speed of light 
along their direction of motion.

Schematically, the before and after SSB scenario is illustrated below:



 -       -       -    -
| H⁺ + iH^- | -> |   0   |
| H + iH⁰ |    | f + H |
 -       -       -    -

Under infinitesinal rotations the vacuum, φ₀, should be symmetric.  
Therefore, 

φ₀ -> exp(iθZ)φ₀ where Z is the rotation.

This means that, 

φ₀ -> (1 + iθZ)φ₀ 

So Zφ₀ must equal 0.

SU(2)_L:

       -     - -     -      -     -
σ₃φ₀ = | 1  0 ||   0   | = -|   0   |  ≠ 0 -> broken 
     | 0 -1 || f + H |    | f + H |
       -     - -     -      -     -

U(1)_Y:

       -    - -     -     -     -
Y_Wφ₀ = | 1 0 ||   0   | = |   0   |  ≠ 0 -> broken 
     | 0 1 || f + H |   | f + H |
       -    - -     -     -     -

U(1)_EM:

                       -    - -     -     
Qφ₀ = (1/2)(σ₃ + Y_W) = | 1 0 ||   0   | = 0 -> unbroken 
                    | 0 0 || f + H |   
                       -    - -     -     

After SSB, the H⁺, H^- and H⁰ components all become Goldstone bosons.  
The 4th component, H, gets a VEV, f.  The 3 Goldstone bosons get 'eaten' 
by combinations of the massless W_μ⁰, W_μ¹, W_μ² and B_μ field bosons in the 
following way:

The negative combination of W_μ¹ and W_μ² 'eats' H⁺ giving the W⁺ boson 
a longitudinal 3rd polarization (i.e. mass).

Likewise, the positive combination of W_μ¹ and W_μ² 'eats' H^- to give the 
W^- boson a mass.

And the negative combination of W_μ⁰ and B_μ 'eats' H⁰ to give the 
Z⁰ boson a mass.

Finally, the leftover combination of W_μ⁰ and B_μ forms the photon.  But 
since there are no more Goldstone bosons to 'eat' the photon remains 
massless.

We could repeat this process with (f + H) to see the involvement of 
the the Higgs boson.  However, the math is more cumbersome and 
therefore is not attempted here.  The end result is the Higgs boson 
component forms a condensate which, as we will see, can interact 
with fermions to also give them mass.

SU(3) ⊗ SU(2) ⊗ U(1) Symmetry Breaking
---------------------------------------

Under a local SU(3) transformation the field transforms as:

φ -> φexp(iθ_i(x)λ_i/2)

where λ_i are the Gell-Mann matrices - the generators of SU(3).  Since 
there are 8 generators there are 8 gauge fields designated by G_μⁱ which 
correspond to the gluons.  

The covariant derivative for a SU(3)⊗SU(2)⊗U(1) gauge invariant Lagrangian 
is:

D_μ = ∂_μ + ig''G_μⁱλ_i/2 + igW_μⁱσ_i/2 + ig'Y_WB_μ

Where g, g' and g'' are coupling constants.

Due to the non-commutativity of the generators of this symmetry, the 
gauge tensor is:

G_μν = ∂_μG_νⁱ - ∂_νG_μⁱ + f^ijkG_μ^jG_ν^k

where f^ijk is the structure constant of the particular gauge group.

Fermionic Mass
--------------

     -     -
φ = |   0   | e^iα
    | f + H |
     -     -
      -        -
φ^† = | 0  f + H | e^-iα
      -        -

The fermions also get their mass from the spontaneous symmetry breaking 
of the Higgs Field.  Fermions include quarks, neutrinos, electrons, muons 
and tau's.

It has been experimentally discovered that in the decay of a neutron to 
a proton, the electron emitted is always left handed.  This turns out to 
be true for any process involving the weak interaction.  For example, 
an interaction involving a W boson and an electron to produce a neutrino,
will only take place if the electron is left handed.  In addition, the 
neutrino produced will always be left handed.  

For any fermion we can write the Dirac equation in terms of right 
and left handed particles as follows (h = c = 1):

i∂ψ_R/∂t + iα_i∂ψ_R/∂x = mψ_L

i∂ψ_L/∂t + iα_i∂ψ_L/∂x = mψ_R

Lets consider the electron.  The mass terms combine two particles, e_R and 
e_L into a single particle - the 'physical electron' which is propagating 
through space.  e_L and e_R swap back and forth.  At a particular point 
in time the particle may be e_L, but if you observe it a moment later, 
the very same particle might manifest itself as e_R.  Weak interactions
with the 'physical electron' occur through its left-handed component 
and not with the right-handed component. 

At this point we need to consider the weak hypercharge associated with 
the weak interaction.  The weak hypercharge, Y_W, relates the electric 
charge, Q, and I₃.  It satisfies the equality:

Q = I₃ + Y_W/2  

Where Q is the charge and I₃ is the third component of the isospin.

∴  Y_W = 2(Q - I₃)

Consider the case for electrons. Y_W for a left-handed electron, e_L, 
is -1 and for a right-handed electron, e_R, is -2.  It is clear that, 
as it stands, plugging these values into the 2 Dirac equations 
results in a charge violation that 'breaks' the coupling between 
the two versions of the electron.  

Now, the Higgs field, H, has a weak hypercharge equal to 1.  

If we rewrite the Dirac Equations as:

i∂ψ_R/∂t + iα_i∂ψ_R/∂x = mH^†ψ_L

i∂ψ_L/∂t + iα_i∂ψ_L/∂x = mHψ_R

Under SSB, the weak hypercharge is now conserved.  In essence, the
Higgs field become a condensate that acts as a source and sink of  
the weak hypercharge.  The addition/subtraction of H (f) can be 
looked at in the following way:
                  
                 /H (provides weak hypercharge)
                /
        rh------g_y
                \
                 \ lh


                 /H^†(removes weak hypercharge)
                /
        lh------g_y
                \
                 \ rh

Now, H = f (the VEV) and we can write the 2 Dirac equations as (f^† = f):

i∂ψ_R/∂t + iα_i∂ψ_R/∂x = g_yfψ_L

i∂ψ_L/∂t - iα_i∂ψ_L/∂x = g_yfψ_R

where g_y is the YUKAWA coupling constant and represents the interaction 
between the Higgs field and a Dirac field.

It is apparent from this the mass term in the original equations 
is equivalent to g_yf.  Therefore, we can write:

     m = g_yf

There is a g_y for every type of fermion. Example, for an electron - 
W boson interaction: 

 g_y ~ 90GeV/0.5Mev ~ 10^-5

The above calculations have assumed the low energy case where we 
ignored the oscillations of the Higgs field and assumed that f is 
a constant.  To include the Higgs boson we would need to rewrite 
the right hand side of the equations as follows:

i∂ψ_R/∂t + iα_i∂ψ_R/∂x = g_y(f + H)ψ_L

i∂ψ_L/∂t + iα_i∂ψ_L/∂x = g_y(f + H)ψ_R

Alternative Derivation
----------------------

We can also derive similar results from the Lagrangian of the Dirac 
equation.  The Dirac field has 4 components:  Positive and negative 
energies and left and right handedness (spin).

The Dirac equation with h = c = 1 is: 

(iγ^μ∂_μ - m)ψ = 0

Solutions to this equation are the DIRAC SPINORS.
                                  _       _
To get the Lagrangian multiply by ψ where ψ = γ⁰ψ^† ... the DIRAC
ADJOINT SPINOR.  Thus,
    _
L = ψ(iγ^μ∂_μ - m)ψ 
_
ψ is interpreted as a creation operator and ψ is interpreted as an
annihilation operator. 

Note: We can prove that this is the Lagrangian by applying the 
Euler-Lagrange equations.  The result yields the original Dirac 
equation.  Also, this is not the Lagrangian for QED. The QED 
Lagrange contains D_μ and the electromagnetic
tensor.

     -  -          -             -           -               -
ψ = | ψ₁ |   ψ^† =  | ψ₁^† ψ₂^† ψ₃^† ψ₄^† |   γ⁰ψ^† = | ψ₁^† ψ₂^† -ψ₃^† -ψ₄^† |
    | ψ₂ |         -             -           -               -
    | ψ₃ |
    | ψ₄ |
     -  -

The purpose of γ⁰ is to make the Lagrangian Lorentz invariant.  
Rewrite the DE as:
 _         _
iψγ^μ∂_μψ - mψψ = 0

The first term in the Lagrangian is the kinetic term. The second term
flips the chirality back and forth.  We can write:

The Dirac spinor, ψ, can be written as a linear combinations of 
left-handed and right-handed states.  Therefore,

ψ = ψ_L + ψ_R and ψ^† = ψ_L^† + ψ_R^†
                             _
Thus, using the relationship ψ = ψ^†γ⁰, we get:
_
ψψ = ψ^†γ⁰ψ

      -      -  -   -  -  - 
   = | ψ^†_L ψ^†_R || 0 1 || ψ_L | 
      -      - | 1 0 || ψ_R |
                -   -  -  - 

   = ψ^†_Lψ_R + ψ^†_Rψ_L

The first term of the Lagrangian can be written as:

L_KE = iψ^†_Lγ^μ∂_μψ_L + iψ^†_Rγ^μ∂_μψ_R

In order to account for the interaction of the fermions with the 
symmetry group SU(2)⊗U(1) and the gauge fields we have to replace 
∂_μ with the gauge covariant derivative, D_μ. Thus,
     
L_KE = iψ^†_Lγ^μD_μψ_L + iψ^†_Rγ^μD_μψ_R

The second term of the Lagrangian can be written as:
  _  
mψψ = mψ^†_Lψ_R + mψ^†_Rψ_L 

However, this as it stands would not be a legal term in the Lagrangian
because it violates weak hypercharge conservation.  As before, we can 
restore the coupling if we introduce the Higgs field, H, as follows:
    _ 
g_yψHψ = g_y(ψ^†_LHψ_R + ψ^†_RH^†ψ_L)
          _
where g_yψHψ is the YUKAWA INTERACTION

Consider the case H = f.  We get:
    _      
g_yψfψ = g_yf(ψ^†_Lψ_R + ψ^†_Rψ_L)

Now consider f + H.  We get:
          _     
g_yψ(f + H)ψ = g_y(ψ^†_L(f + H)ψ_R + ψ^†_R(f + H)ψ_L)

So the RHS becomes:
      
g_yf(ψ^†_Lψ_R + ψ^†_Rψ_L) + g_yH(ψ^†_Lψ_R + ψ^†_Rψ_L) + the hermitian conjugate

The first term is the fermion mass.  The second term describes a
process whereby a fermion is absorbed and re-emitted alongside a
Higgs boson.

The Complete Electroweak Lagrangian
-----------------------------------

Before SSB:

L_EW = L_Higgs + L_Gauge + L_Dirac + L_Yukawa

  = D_μφD^μφ^† - V(φ) - F_μνF^μν - G_μνⁱG^μν_i + iψ_Lγ^μD_μψ_L + iψ_Rγ^μD_μψ_R

  + g_y(ψ_Lφψ_R + ψ_Rφ^†ψ_L)
 
Majorana Mass
-------------

The kind of fermion mass that we discussed above is called a Dirac mass. 
This is a type of mass that connects two different particles, e_R and e_L.
It is also possible to have a mass that connects two of the same kind of 
particle, this is called a MAJORANA mass. 

This type of mass is forbidden for particles that have any type of charge.
There is, however, one type of matter particle in the Standard Model which 
does not carry any charge - the neutrino.  Majorana masses mix neutrinos 
with anti-neutrinos so that the “physical neutrino” is its own antiparticle.